1115. Counting Nodes in a BST (30)

时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following PRoperties:

The left subtree of a node contains only nodes with keys less than or equal to the node's key. The right subtree of a node contains only nodes with keys greater than the node's key. Both the left and right subtrees must also be binary search trees.

Insert a sequence of numbers into an initially empty binary search tree. Then you are supposed to count the total number of nodes in the lowest 2 levels of the resulting tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=1000) which is the size of the input sequence. Then given in the next line are the N integers in [-1000 1000] which are supposed to be inserted into an initially empty binary search tree.

Output Specification:

For each case, print in one line the numbers of nodes in the lowest 2 levels of the resulting tree in the format:

n1 + n2 = n

where n1 is the number of nodes in the lowest level, n2 is that of the level above, and n is the sum.

Sample Input:

9
25 30 42 16 20 20 35 -5 28

Sample Output:
2 + 4 = 6




题意：计算二叉搜索树最后两层的结点数

#include <cstdio>
#include <cstring>
#include <iostream>
#include <stdlib.h>
using namespace std;

struct node
{
    int v;
    node *left;
    node *right;
    node(int v):v(v),left(NULL),right(NULL){}
};

int n, t, a[1010] = {0};

void build(node* &root, int v, int step)
{
    if(root == NULL)
    {
        root = new node(v);
        a[step]++;
        return;
    }
    if(v <= root->v) build(root->left, v, step+1);
    else build(root->right, v, step+1);
}

int main()
{
    node* root = NULL;
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
    {
        scanf("%d", &t);
        build(root, t, 1);
    }
    int last = 0;
    for(int i = 1; a[i]!=0; i++)
        last = i;
    printf("%d + %d = %d\n", a[last], a[last-1], a[last] + a[last-1]);
    return 0;
}